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Section 5.1 #54: Determine if the function is one-to-one. If it is, find a formula for the inverse function: f(x)=(x+5)3.
Solution: First, the graph of this function is obtained by starting with the graph of y=x3 and then shifting it to the left by 5 units:
,
which passes the "horizontal line test" and hence is one-to-one. So finding the inverse is possible. We have y=(x+5)3. Solve for x by first taking the cube root to get 3y=3(x+5)3. So, 3y=x+5. Therefore 3y5=x. This means that f1(y)=3y5.
(note: writing f1(x)=3x5 is also ok -- you would arrive here if you do the "switch x and y" thing)

Section 5.2 #34: Sketch the graph of y=2x13.
Solution: This is the graph of y=2x shifted to the right by 1 and then shifted down by 3. Therefore we get:



Section 5.3 #10: Find log3(9).
Solution: The key is to recall that log3(3x)=x, and so we want to rewrite 9 as a power of 3. This yields log3(9)=log3(32)=2.

Section 5.3 #33: Find ln(e).
Solution: Recall that ln is simply loge, where e is the special number e=2.71. Therefore we want to express e as a power of e and then take the logarithm: ln(e)=ln(e12)=12.

Section 5.3 #46: Convert to an exponential equation: t=log4(7).
Solution: We would like the logarithm on the right to disappear. Recall that 4log4(x)=x, and so we plug both sides of the equation into 4x to get 4t=4log4(7) yielding 4t=7.

Section 5.3 #90: Sketch the graph of y=ln(x+1).
Solution: This is the graph of y=ln(x) shifted to the left by 1 unit. Therefore,